13.8: Solution Stoichiometry

As we learned in Chapter 7, double replacement reactions involve the reaction between ionic compounds in solution and, in the course of the reaction, the ions in the two reacting compounds are “switched” (they replace each other). Because these reactions occur in aqueous solution, we can use the concept of molarity to directly calculate the number of moles of reactants or products that will be formed, and hence their amounts (i.e. volume of solutions or mass of precipitates).

As an example, lead(II) nitrate and sodium chloride react to form sodium nitrate and the insoluble compound, lead(II) chloride.

In the reaction shown above, if we mixed 0.123 L of a 1.00 M solution of \(\ce{NaCl}\) with 1.50 M solution of \(\ce{Pb(NO3)2}\), we could calculate the volume of \(\ce{Pb(NO3)2}\) solution needed to completely precipitate the \(\ce{Pb^{2+}}\) ions.

First, we must examine the reaction stoichiometry in the balanced reaction (Equation \ref{EQ1}). In this reaction, one mole of \(\ce{Pb(NO3)2}\) reacts with two moles of \(\ce{NaCl}\) to give one mole of \(\ce{PbCl2}\) precipitate. Thus, the concept map utilizing the stoichiometric ratios is:

This volume make intuitive sense for two reasons: (1) the number of moles of \(\ce{Pb(NO3)2}\) required is half of the number of moles of \(\ce{NaCl}\) based off of the stoichiometry in the balanced reaction (Equation \ref{EQ1}) and (2) the concentration of \(\ce{Pb(NO3)2}\) solution is 50% greater than the \(\ce{NaCl}\) solution, so less volume is needed.

Example \(\PageIndex{1}\)

What volume (in L) of 0.500 M sodium sulfate will react with 275 mL of 0.250 M barium chloride to completely precipitate all \(\ce{Ba^{2+}}\) in the solution?

Solution

Steps for Problem Solving

Example \(\PageIndex{1}\)

Identify the "given"information and what the problem is asking you to "find."

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